Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).
Differential equations
Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m
Matrices & determinants
2×2 determinant
det = ad − bc
2×2 inverse
A⁻¹ = (1/det)·[d −b; −c a]
Product
(AB)ij = Σk AikBkj
Eigenvalues
det(A − λI) = 0
Cramer’s rule
xi = det(Ai)/det(A)
Vectors: dot & cross product
Magnitude
|a| = √(ax² + ay² + az²)
Unit vector
û = a/|a|
Components from an angle
Fx = F·cos θ, Fy = F·sin θ
Dot product
a·b = axbx + ayby + azbz = |a||b|cos θ
Scalar projection of a onto b
a·b/|b|
Work
W = F·d
Cross product
a × b = (aybz − azby, azbx − axbz, axby − aybx)
Areas
parallelogram |a × b|, triangle ½|a × b|
Moment about O
MO = r × F, r = A − O
Volume of a parallelepiped
V = |a·(b × c)|
Method
Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
Use the cross product for perpendicular vectors, areas and moments: write the three components carefully, keeping the order a × b.
Probability
Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k
Descriptive statistics
Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ
Normal distribution
z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ
Method
Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
For a percentile, work backwards: find z for the area, then x = μ + z·σ.
2Practice problems
1.LimitsMedium
Find the limit L = limx→∞ (4x3 + 5x − 8)/(x3 + 4x + 8).
f(x) = (4x3 + 5x − 8)/(x3 + 4x + 8)x → ∞
L =
2.LimitsHard
Find the limit L = limx→0 (ex − 1 − x)/x².
f(x) = (ex − 1 − x)/x²x → 0
L =
3.DerivativesMedium
Let f(x) = x·sin(3x), with x in radians. Find f′(0.5).
f(x) = x·sin(3x)x = 0.5 rad
f′(0.5) =
4.DerivativesHard
The tangent line to f(x) = x2 − 5x − 1 at x = 1 crosses the y-axis at y = b. Find b.
f(x) = x2 − 5x − 1x = 1
b =
5.IntegralsMedium
Evaluate I = ∫₀π/3 x·sin x dx (x in radians).
∫₀π/3 x·sin x dx
I =
6.IntegralsHard
Evaluate I = ∫₀2 x²·eˣ dx.
∫₀2 x²eˣ dx
I =
7.Complex numbersMedium
The diagram shows z = 1 + j4. With w = 0.5∠90°, click the point that represents z/w.
z = 1 + j4w = 0.5∠90°
1A (2 + j8)
2B (8 − j2)
3C (4 − j)
4D (−8 + j2)
8.Complex numbersHard
The roots of zⁿ = w with n = 5 and w = 243∠180° are zk = R1/n∠((φ + 360°·k)/n), k = 0, 1, …, 4. Find the argument of z₀ in degrees, between −180° and 180°.
w = 243∠180°n = 5k = 0
θ₀ = °
9.Differential equationsMedium
A cup of coffee at 89 °C is left in a room at 21 °C. After 8 min its temperature is 74 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 89 °CTa = 21 °CT(8) = 74 °C
k = min⁻¹
10.Differential equationsHard
A tank holds 100 L of water in which 5 kg of salt is dissolved. Brine with 0.25 kg of salt per litre flows in at 2 L/min, and the well-mixed solution flows out at the same rate. How much salt m is in the tank after 15 min?
V = 100 Lm0 = 5 kgc = 0.25 kg/Lr = 2 L/min
m = kg
11.Matrices & determinantsMedium
Find the determinant det A of the 3×3 matrix A shown.
A = [−3 4 −3; 5 0 1; −4 −3 −5]
det A =
12.Matrices & determinantsHard
Solve the system −3x + 4y = 5 and −2x + 3y = 8. Find x.
−3x + 4y = 5−2x + 3y = 8
x =
13.Vectors: dot & cross productMedium
Find the dot product a·b of a = (−4, −7, −2) and b = (−1, 3, 5).
a = (−4, −7, −2)b = (−1, 3, 5)
a·b =
14.Vectors: dot & cross productHard
Find the area A of the parallelogram spanned by a = (−5, −3, 4) and b = (−4, 0, −6).
a = (−5, −3, 4)b = (−4, 0, −6)
A =
15.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.1, 0.2, 0.35, 0.05
Var(X) =
16.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 1 correct answers?
n = 9p = 0.3k = 1
P =
17.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 14, 48, 60, 27, 36, 16.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is below 120?
μ = 100σ = 10x = 120
P =
19.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 171 and 179.75 cm? Give P as a decimal.
μ = 178σ = 7x1 = 171x2 = 179.75
P =
20.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 75% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g75%
x = g
3Solutions
1.Limits
Answer: L = 4
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
Divide by the highest power
(4 + 5/x2 + (−8)/x3)/(1 + 4/x2 + 8/x3)
Let x → ∞
L = 4/1 = 4
Answer
L = 4
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 0.5
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
First application
(1·ex − 1)/(2x) → 0/0
Second application
1·ex/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(0.5) = 1.104
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Product and chain rule
f′(x) = sin(3x) + 3x·cos(3x)
Substitute
f′(0.5) = 0.9975 + 1.5·0.07074 = 1.104
Answer
f′(0.5) = 1.104
4.Derivatives
Answer: b = −2
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Point and slope
f(1) = −5
f′(x) = 2x − 5 → f′(1) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = −5 − (−3)·1 = −2
Answer
b = −2
5.Integrals
Answer: I = 0.3424
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
Choose u and dv
u = x, du = dx; dv = sin x dx, v = −cos x
Antiderivative
∫x sin x dx = −x cos x + ∫cos x dx = −x cos x + sin x
Limits
(−π/3·cos π/3 + sin π/3) − (0 + 0) = 0.3424
Answer
I = 0.3424
6.Integrals
Answer: I = 12.78
Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.
First round
∫x²eˣ dx = x²eˣ − ∫2x eˣ dx
Second round
∫2x eˣ dx = 2x eˣ − 2eˣ
∫x²eˣ dx = eˣ(x² − 2x + 2)
Limits
7.389·(2) − 1·2 = 12.78
Answer
I = 12.78
7.Complex numbers
Answer: B (8 − j2)
Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.
Effect of w
|z/w| = |z|/0.5, arg(z/w) = arg z − 90°
The point
z/w = 8 − j2 → B
Answer
B (8 − j2)
8.Complex numbers
Answer: θ₀ = 36 °
De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.
Modulus of the roots
2431/5 = 3
Argument of the root
θ₀ = (180° + 360°·0)/5 = 36°
Argument
θ₀ = 36°
Answer
θ₀ = 36°
The 5 roots lie on the circle of radius 3, 72° apart.
9.Differential equations
Answer: k = 0.03115 min⁻¹
First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.
Ratio of temperature differences
e−8k = 53/68
Solve for k
k = ln(68/53)/8 = 0.03115 min⁻¹
Answer
k = 0.03115 min⁻¹
10.Differential equations
Answer: m = 10.18 kg
First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.
Set up the equation
dm/dt = 2·0.25 − (2/100)·m = 0.5 − 0.02·m
Solve
The integrating factor is ert/V; the solution approaches the equilibrium c·V.
m(t) = 25 + (−20)·e−t/50
Substitute t
m(15) = 25 + (−20)·0.7408 = 10.18 kg
Answer
m = 10.18 kg
11.Matrices & determinants
Answer: det A = 120
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
2×2 minors of the first row
M₁₁ = 3, M₁₂ = −21, M₁₃ = −15
Cofactor expansion
det A = (−3)·3 − 4·(−21) + (−3)·(−15) = 120
Answer
det A = 120
12.Matrices & determinants
Answer: x = 17
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
Determinant of the coefficient matrix
det A = (−3)·3 − 4·(−2) = −1
Replace the x column
det Aₓ = 5·3 − 4·8 = −17
x = −17/(−1) = 17
Answer
x = 17
y = 14
13.Vectors: dot & cross product
Answer: a·b = −27
Dot product:a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.
Negative: the angle between the vectors is more than 90°.
14.Vectors: dot & cross product
Answer: A = 50.83
Cross product:a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.
Cross product
(a × b)x = aybz − azby = (−3)·(−6) − 4·0 = 18
(a × b)y = azbx − axbz = 4·(−4) − (−5)·(−6) = −46
(a × b)z = axby − aybx = (−5)·0 − (−3)·(−4) = −12
Area
A = |a × b| = √(18² + (−46)² + (−12)²) = 50.83
Answer
A = 50.83
15.Probability
Answer: Var(X) = 1.787
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Binomial formula
P(X = 1) = 9·0.31·0.78 = 9·0.3·0.05765 = 0.1556
Answer
P = 0.1556
17.Descriptive statistics
Answer: s = 18.15
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
19.Normal distribution
Answer: P = 0.4401
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (171 − 178)/7 = −1
z₂ = (179.75 − 178)/7 = 0.25
Area between
P = Φ(0.25) − Φ(−1) = 0.5987 − 0.1587 = 0.4401
Answer
P = 0.4401
20.Normal distribution
Answer: x = 246.7 g
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.